Projective Transformation for two distinct vertices and a linear objective function The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Projective and affine conic classificationVisualizing projective closures - is it okay to just think of the affine case?Optimizing over intersection of polytopes inside polytopeAre two halves of a convex polytope themselves convex?Extreme points of intersection of the orthant (quadrant) with an Hyperplane in finite dimension vectorial spaceUsing co-planarity to determine the points inside a triangleDifferences/ Similarities between different kinds of geometries.Point of intersection of lines in $mathbbRP^2$Projective classification of quadric from affine classificationIdentifying the Plane at Infinity in the World Necessitates Determining the Affine Geometry of the World?
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Projective Transformation for two distinct vertices and a linear objective function
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Projective and affine conic classificationVisualizing projective closures - is it okay to just think of the affine case?Optimizing over intersection of polytopes inside polytopeAre two halves of a convex polytope themselves convex?Extreme points of intersection of the orthant (quadrant) with an Hyperplane in finite dimension vectorial spaceUsing co-planarity to determine the points inside a triangleDifferences/ Similarities between different kinds of geometries.Point of intersection of lines in $mathbbRP^2$Projective classification of quadric from affine classificationIdentifying the Plane at Infinity in the World Necessitates Determining the Affine Geometry of the World?
$begingroup$
I am trying to understand how to prove the following, which seems to be a quite useful insight in terms of linear optimization. Unfortunately, I have a hard time with projective geometry. I'd greatly appreciate if someone could explain me how this can be done.
Suppose we have an $n$-dimensional polytope $P subseteq mathbbR^n$
and two distinct vertices $u$ and $v$. I want to show that there
exists a projective transformation $P rightarrow P'$ such that the
transformed vertices $u'$ and $v'$ have the smallest, respectively the
largest, $x_n$ coordinate among all vertices of $P'$.
Thanks in advance!
UPDATE: I think that this statement is not true if we replace "projective transformation" with "affine transformation". It might be true if we restrict to neighboring vertices $u$ and $v$ but I am not quite sure about that either. So if we suppose that this doesn't work in affine geometry, but it does in projective geometry, then I'd reckon that we somehow have to exploit the ability to map facets to the infinity hyperplane..
projective-geometry projective-space polytopes discrete-geometry
$endgroup$
This question has an open bounty worth +50
reputation from Doc ending ending at 2019-04-15 13:19:50Z">in 2 days.
This question has not received enough attention.
add a comment |
$begingroup$
I am trying to understand how to prove the following, which seems to be a quite useful insight in terms of linear optimization. Unfortunately, I have a hard time with projective geometry. I'd greatly appreciate if someone could explain me how this can be done.
Suppose we have an $n$-dimensional polytope $P subseteq mathbbR^n$
and two distinct vertices $u$ and $v$. I want to show that there
exists a projective transformation $P rightarrow P'$ such that the
transformed vertices $u'$ and $v'$ have the smallest, respectively the
largest, $x_n$ coordinate among all vertices of $P'$.
Thanks in advance!
UPDATE: I think that this statement is not true if we replace "projective transformation" with "affine transformation". It might be true if we restrict to neighboring vertices $u$ and $v$ but I am not quite sure about that either. So if we suppose that this doesn't work in affine geometry, but it does in projective geometry, then I'd reckon that we somehow have to exploit the ability to map facets to the infinity hyperplane..
projective-geometry projective-space polytopes discrete-geometry
$endgroup$
This question has an open bounty worth +50
reputation from Doc ending ending at 2019-04-15 13:19:50Z">in 2 days.
This question has not received enough attention.
add a comment |
$begingroup$
I am trying to understand how to prove the following, which seems to be a quite useful insight in terms of linear optimization. Unfortunately, I have a hard time with projective geometry. I'd greatly appreciate if someone could explain me how this can be done.
Suppose we have an $n$-dimensional polytope $P subseteq mathbbR^n$
and two distinct vertices $u$ and $v$. I want to show that there
exists a projective transformation $P rightarrow P'$ such that the
transformed vertices $u'$ and $v'$ have the smallest, respectively the
largest, $x_n$ coordinate among all vertices of $P'$.
Thanks in advance!
UPDATE: I think that this statement is not true if we replace "projective transformation" with "affine transformation". It might be true if we restrict to neighboring vertices $u$ and $v$ but I am not quite sure about that either. So if we suppose that this doesn't work in affine geometry, but it does in projective geometry, then I'd reckon that we somehow have to exploit the ability to map facets to the infinity hyperplane..
projective-geometry projective-space polytopes discrete-geometry
$endgroup$
I am trying to understand how to prove the following, which seems to be a quite useful insight in terms of linear optimization. Unfortunately, I have a hard time with projective geometry. I'd greatly appreciate if someone could explain me how this can be done.
Suppose we have an $n$-dimensional polytope $P subseteq mathbbR^n$
and two distinct vertices $u$ and $v$. I want to show that there
exists a projective transformation $P rightarrow P'$ such that the
transformed vertices $u'$ and $v'$ have the smallest, respectively the
largest, $x_n$ coordinate among all vertices of $P'$.
Thanks in advance!
UPDATE: I think that this statement is not true if we replace "projective transformation" with "affine transformation". It might be true if we restrict to neighboring vertices $u$ and $v$ but I am not quite sure about that either. So if we suppose that this doesn't work in affine geometry, but it does in projective geometry, then I'd reckon that we somehow have to exploit the ability to map facets to the infinity hyperplane..
projective-geometry projective-space polytopes discrete-geometry
projective-geometry projective-space polytopes discrete-geometry
edited Apr 8 at 12:13
Doc
asked Apr 6 at 12:42
DocDoc
322114
322114
This question has an open bounty worth +50
reputation from Doc ending ending at 2019-04-15 13:19:50Z">in 2 days.
This question has not received enough attention.
This question has an open bounty worth +50
reputation from Doc ending ending at 2019-04-15 13:19:50Z">in 2 days.
This question has not received enough attention.
add a comment |
add a comment |
1 Answer
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$begingroup$
I guess it is not very rigorous to say about projective transformations of subsets of an affine space, beacuse they should deal with points at infinity, which formally do not belong to the space. So I have the following hand-waving idea. Since the polytope $P$ is convex, there exist hyperplanes $H_uni u$ and $H_vni v$ supporting $P$. If they are parallel then we draw $x$-axis perpendicularly to them. Otherwise we move the intersection $H_ucap H_v$ to infinity by a projective transformation, which will make the hyperplanes $H_u$ and $H_v$ parallel.
$endgroup$
1
$begingroup$
Thanks for the idea. I tried this using a particular polytope example and computed the projection etc, to get a better idea of projective transformation. What I still not have figured out. Lets say I use a projective transformation to move said intersection to infinity. What guarantees me that this projection is actually admissible for P, hence leaving me with a new proper polytope after transforming.
$endgroup$
– Doc
yesterday
$begingroup$
@Doc Let $(x, n_u)=c_u$ and $(x, n_v)=c_v$ be the equations of the hyperplanes $H_u$ and $H_v$, respectively such that $(y,n_u)le c_u$ and $(y,n_v)le c_v$ for each $yin P$. It is easy to check that a hyperplane $H$ determined by a equation $(x, n_u+n_v)=c_u+c_v$ contains $H_ucap H_v$ and is disjoint from $P$. I guess when we make $H$ the infinity hyperplane, $P$ will be transformed to a proper polytope.
$endgroup$
– Alex Ravsky
18 hours ago
add a comment |
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$begingroup$
I guess it is not very rigorous to say about projective transformations of subsets of an affine space, beacuse they should deal with points at infinity, which formally do not belong to the space. So I have the following hand-waving idea. Since the polytope $P$ is convex, there exist hyperplanes $H_uni u$ and $H_vni v$ supporting $P$. If they are parallel then we draw $x$-axis perpendicularly to them. Otherwise we move the intersection $H_ucap H_v$ to infinity by a projective transformation, which will make the hyperplanes $H_u$ and $H_v$ parallel.
$endgroup$
1
$begingroup$
Thanks for the idea. I tried this using a particular polytope example and computed the projection etc, to get a better idea of projective transformation. What I still not have figured out. Lets say I use a projective transformation to move said intersection to infinity. What guarantees me that this projection is actually admissible for P, hence leaving me with a new proper polytope after transforming.
$endgroup$
– Doc
yesterday
$begingroup$
@Doc Let $(x, n_u)=c_u$ and $(x, n_v)=c_v$ be the equations of the hyperplanes $H_u$ and $H_v$, respectively such that $(y,n_u)le c_u$ and $(y,n_v)le c_v$ for each $yin P$. It is easy to check that a hyperplane $H$ determined by a equation $(x, n_u+n_v)=c_u+c_v$ contains $H_ucap H_v$ and is disjoint from $P$. I guess when we make $H$ the infinity hyperplane, $P$ will be transformed to a proper polytope.
$endgroup$
– Alex Ravsky
18 hours ago
add a comment |
$begingroup$
I guess it is not very rigorous to say about projective transformations of subsets of an affine space, beacuse they should deal with points at infinity, which formally do not belong to the space. So I have the following hand-waving idea. Since the polytope $P$ is convex, there exist hyperplanes $H_uni u$ and $H_vni v$ supporting $P$. If they are parallel then we draw $x$-axis perpendicularly to them. Otherwise we move the intersection $H_ucap H_v$ to infinity by a projective transformation, which will make the hyperplanes $H_u$ and $H_v$ parallel.
$endgroup$
1
$begingroup$
Thanks for the idea. I tried this using a particular polytope example and computed the projection etc, to get a better idea of projective transformation. What I still not have figured out. Lets say I use a projective transformation to move said intersection to infinity. What guarantees me that this projection is actually admissible for P, hence leaving me with a new proper polytope after transforming.
$endgroup$
– Doc
yesterday
$begingroup$
@Doc Let $(x, n_u)=c_u$ and $(x, n_v)=c_v$ be the equations of the hyperplanes $H_u$ and $H_v$, respectively such that $(y,n_u)le c_u$ and $(y,n_v)le c_v$ for each $yin P$. It is easy to check that a hyperplane $H$ determined by a equation $(x, n_u+n_v)=c_u+c_v$ contains $H_ucap H_v$ and is disjoint from $P$. I guess when we make $H$ the infinity hyperplane, $P$ will be transformed to a proper polytope.
$endgroup$
– Alex Ravsky
18 hours ago
add a comment |
$begingroup$
I guess it is not very rigorous to say about projective transformations of subsets of an affine space, beacuse they should deal with points at infinity, which formally do not belong to the space. So I have the following hand-waving idea. Since the polytope $P$ is convex, there exist hyperplanes $H_uni u$ and $H_vni v$ supporting $P$. If they are parallel then we draw $x$-axis perpendicularly to them. Otherwise we move the intersection $H_ucap H_v$ to infinity by a projective transformation, which will make the hyperplanes $H_u$ and $H_v$ parallel.
$endgroup$
I guess it is not very rigorous to say about projective transformations of subsets of an affine space, beacuse they should deal with points at infinity, which formally do not belong to the space. So I have the following hand-waving idea. Since the polytope $P$ is convex, there exist hyperplanes $H_uni u$ and $H_vni v$ supporting $P$. If they are parallel then we draw $x$-axis perpendicularly to them. Otherwise we move the intersection $H_ucap H_v$ to infinity by a projective transformation, which will make the hyperplanes $H_u$ and $H_v$ parallel.
answered 2 days ago
Alex RavskyAlex Ravsky
43.2k32583
43.2k32583
1
$begingroup$
Thanks for the idea. I tried this using a particular polytope example and computed the projection etc, to get a better idea of projective transformation. What I still not have figured out. Lets say I use a projective transformation to move said intersection to infinity. What guarantees me that this projection is actually admissible for P, hence leaving me with a new proper polytope after transforming.
$endgroup$
– Doc
yesterday
$begingroup$
@Doc Let $(x, n_u)=c_u$ and $(x, n_v)=c_v$ be the equations of the hyperplanes $H_u$ and $H_v$, respectively such that $(y,n_u)le c_u$ and $(y,n_v)le c_v$ for each $yin P$. It is easy to check that a hyperplane $H$ determined by a equation $(x, n_u+n_v)=c_u+c_v$ contains $H_ucap H_v$ and is disjoint from $P$. I guess when we make $H$ the infinity hyperplane, $P$ will be transformed to a proper polytope.
$endgroup$
– Alex Ravsky
18 hours ago
add a comment |
1
$begingroup$
Thanks for the idea. I tried this using a particular polytope example and computed the projection etc, to get a better idea of projective transformation. What I still not have figured out. Lets say I use a projective transformation to move said intersection to infinity. What guarantees me that this projection is actually admissible for P, hence leaving me with a new proper polytope after transforming.
$endgroup$
– Doc
yesterday
$begingroup$
@Doc Let $(x, n_u)=c_u$ and $(x, n_v)=c_v$ be the equations of the hyperplanes $H_u$ and $H_v$, respectively such that $(y,n_u)le c_u$ and $(y,n_v)le c_v$ for each $yin P$. It is easy to check that a hyperplane $H$ determined by a equation $(x, n_u+n_v)=c_u+c_v$ contains $H_ucap H_v$ and is disjoint from $P$. I guess when we make $H$ the infinity hyperplane, $P$ will be transformed to a proper polytope.
$endgroup$
– Alex Ravsky
18 hours ago
1
1
$begingroup$
Thanks for the idea. I tried this using a particular polytope example and computed the projection etc, to get a better idea of projective transformation. What I still not have figured out. Lets say I use a projective transformation to move said intersection to infinity. What guarantees me that this projection is actually admissible for P, hence leaving me with a new proper polytope after transforming.
$endgroup$
– Doc
yesterday
$begingroup$
Thanks for the idea. I tried this using a particular polytope example and computed the projection etc, to get a better idea of projective transformation. What I still not have figured out. Lets say I use a projective transformation to move said intersection to infinity. What guarantees me that this projection is actually admissible for P, hence leaving me with a new proper polytope after transforming.
$endgroup$
– Doc
yesterday
$begingroup$
@Doc Let $(x, n_u)=c_u$ and $(x, n_v)=c_v$ be the equations of the hyperplanes $H_u$ and $H_v$, respectively such that $(y,n_u)le c_u$ and $(y,n_v)le c_v$ for each $yin P$. It is easy to check that a hyperplane $H$ determined by a equation $(x, n_u+n_v)=c_u+c_v$ contains $H_ucap H_v$ and is disjoint from $P$. I guess when we make $H$ the infinity hyperplane, $P$ will be transformed to a proper polytope.
$endgroup$
– Alex Ravsky
18 hours ago
$begingroup$
@Doc Let $(x, n_u)=c_u$ and $(x, n_v)=c_v$ be the equations of the hyperplanes $H_u$ and $H_v$, respectively such that $(y,n_u)le c_u$ and $(y,n_v)le c_v$ for each $yin P$. It is easy to check that a hyperplane $H$ determined by a equation $(x, n_u+n_v)=c_u+c_v$ contains $H_ucap H_v$ and is disjoint from $P$. I guess when we make $H$ the infinity hyperplane, $P$ will be transformed to a proper polytope.
$endgroup$
– Alex Ravsky
18 hours ago
add a comment |
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